help with algebra problem?

x^-2/3 or 1/(3x^2)

Factor out the 3/9, which simplifies to 1/3, and multiply by x^-4/x^-2, which reduces to x^-2 (just subtract the exponents when dividing values with the same base)

(3x^-4)/(9x^-2)
= (3/9)(x^-4/x^-2)
= (1/3)[x^(-4 + 2)]
= (1/3)(x^-2)
= (1/3)(1/x^2)
= 1/(3x^2)

1/3^2

3x^-4/9x^-2 = x^-4 * x^2 / 3 = x^-2/3 = 1/(3x^2)

3 will go in 9 and for powers of x use (a^m)/(a^n) = a^(m-n).
You must have a list of relevant formulae before starting to solve the problems.